Theorem eujust 2021

Description: A soundness justification theorem for df-eu 2022, showing that the definition is equivalent to itself with its dummy variable renamed. Note that 𝑦 and 𝑧 needn't be distinct variables. (Contributed by NM, 11-Mar-2010.) (Proof shortened by Andrew Salmon, 9-Jul-2011.)

Assertion

Ref	Expression
eujust	⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ ∃𝑧∀𝑥(𝜑 ↔ 𝑥 = 𝑧))

Distinct variable groups:   𝑥,𝑦   𝑥,𝑧   𝜑,𝑦   𝜑,𝑧

Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem eujust

Dummy variable 𝑤 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		equequ2 1706	. . . . 5 ⊢ (𝑦 = 𝑤 → (𝑥 = 𝑦 ↔ 𝑥 = 𝑤))
2	1	bibi2d 231	. . . 4 ⊢ (𝑦 = 𝑤 → ((𝜑 ↔ 𝑥 = 𝑦) ↔ (𝜑 ↔ 𝑥 = 𝑤)))
3	2	albidv 1817	. . 3 ⊢ (𝑦 = 𝑤 → (∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑤)))
4	3	cbvexv 1911	. 2 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ ∃𝑤∀𝑥(𝜑 ↔ 𝑥 = 𝑤))
5		equequ2 1706	. . . . 5 ⊢ (𝑤 = 𝑧 → (𝑥 = 𝑤 ↔ 𝑥 = 𝑧))
6	5	bibi2d 231	. . . 4 ⊢ (𝑤 = 𝑧 → ((𝜑 ↔ 𝑥 = 𝑤) ↔ (𝜑 ↔ 𝑥 = 𝑧)))
7	6	albidv 1817	. . 3 ⊢ (𝑤 = 𝑧 → (∀𝑥(𝜑 ↔ 𝑥 = 𝑤) ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑧)))
8	7	cbvexv 1911	. 2 ⊢ (∃𝑤∀𝑥(𝜑 ↔ 𝑥 = 𝑤) ↔ ∃𝑧∀𝑥(𝜑 ↔ 𝑥 = 𝑧))
9	4, 8	bitri 183	1 ⊢ (∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦) ↔ ∃𝑧∀𝑥(𝜑 ↔ 𝑥 = 𝑧))

Syntax hints:   ↔ wb 104  ∀wal 1346   = wceq 1348  ∃wex 1485

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527

This theorem depends on definitions:  df-bi 116

This theorem is referenced by: (None)